/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 The voters in the midterm 2006 e... [FREE SOLUTION] | 91Ó°ÊÓ

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The voters in the midterm 2006 elections showed that Democrat and Republican differences extend beyond matters of opinion, and actually include the way they see the world. \({ }^{12}\) The three most important issues mentioned by Democrats and Republicans are listed below. $$\begin{array}{llll} & \text { 1st } & \text { 2nd } & \text { 3rd } \\\\\hline \text { Republicans } & \text { Terrorism } & \text { Economy } & \text { lraq war } \\\\(n=995) & 42 \% & 41 \% & 37 \% \\\\\text { Democrats } & \text { Iraq war } & \text { Economy } & \text { Health care } \\\\(n=1094) & 60 \% & 44 \% & 44 \%\end{array}$$ Use a large-sample estimation procedure to compare the proportions of Republicans and Democrats who mentioned the economy as an important issue in the elections. Explain your conclusions.

Short Answer

Expert verified
Based on the z-test for proportions, the confidence interval for the difference in proportions between Republicans and Democrats mentioning the economy as an important issue in the midterm 2006 elections is (-0.03, 0.01). Since this interval includes zero, we can conclude that there is no significant difference between the two groups at a 95% confidence level.

Step by step solution

01

Identify the sample proportions

In the given data, 41% of Republicans (n=995) and 44% of Democrats (n=1094) mentioned the economy as an important issue. We can represent these as sample proportions denoted as \(p_{1}\) and \(p_{2}\) for Republicans and Democrats respectively: $$p_1 = 0.41$$ $$p_2 = 0.44$$
02

Calculate the pooled proportion

The pooled proportion is the combined proportion of both groups mentioning the economy. We can calculate this value with the formula: $$p = \frac{p_{1}n_{1} + p_{2}n_{2}}{n_{1}+n_{2}}$$ Plugging in values: $$p = \frac{(0.41)(995) + (0.44)(1094)}{995+1094}$$ $$p \approx 0.4259$$
03

Calculate the z-score

The z-score, denoted as z, is computed using the following formula: $$z = \frac{(p_1 - p_2) - 0}{\sqrt{\frac{p(1-p)}{n_1} + \frac{p(1-p)}{n_2}}}$$ Plugging in the values: $$z = \frac{(0.41 - 0.44) - 0}{\sqrt{\frac{0.4259(1-0.4259)}{995} + \frac{0.4259(1-0.4259)}{1094}}}$$ $$z \approx -1.996$$
04

Determine the margin of error and confidence interval

At a 95% confidence level, the z-critical value is approximately 1.96 (from the z-distribution table). The margin of error can be calculated using the following formula: $$MOE = z_{\text{critical}} \times \sqrt{\frac{p(1-p)}{n_1} + \frac{p(1-p)}{n_2}}$$ Plugging in the values: $$MOE = 1.96 \times \sqrt{\frac{0.4259(1-0.4259)}{995} + \frac{0.4259(1-0.4259)}{1094}}$$ $$MOE \approx 0.029$$ Now, we calculate the 95% confidence interval (difference in proportions \(\pm\) margin of error): $$( -0.03, 0.01)$$
05

Draw conclusions

Since the confidence interval contains zero (0), we can conclude that there is no significant difference in proportions between Republicans and Democrats mentioning the economy as an important issue in the midterm 2006 elections at a 95% confidence level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
In statistics, a confidence interval is a range of values that's used to estimate the true value of an unknown population parameter.
A confidence interval is calculated from the statistics of the observed data, and provides a range of values that are likely to contain the parameter of interest.
For example, in this exercise, the confidence interval addresses the difference in proportions of Republicans and Democrats who consider the economy an important issue.
Typically, a 95% confidence interval is used, which implies that if we were to take 100 different samples and compute a confidence interval for each, we expect about 95 of those intervals to contain the true difference in proportions.
The formula for a simple confidence interval for the difference in proportions is:
  • Difference in proportions \((p_1 - p_2)\) ± Margin of Error (MOE)
The MOE is derived from the critical value of the z-score, which quantifies the level of confidence, multiplied by the standard error of the difference.
Z-score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values.
It tells us how many standard deviations an element is from the mean.
In hypothesis testing, a Z-score is used to determine whether to reject the null hypothesis.
For comparing proportions, like in our exercise, the Z-score helps us understand how different our observed sample proportions are from what we’d expect if there were no differences.
The formula used in this exercise is:
  • \(z = \frac{(p_1 - p_2)}{\sqrt{\frac{p(1-p)}{n_1} + \frac{p(1-p)}{n_2}}}\)
The numerator \(p_1 - p_2\) is the difference between the two sample proportions, and the denominator is the standard error derived from the pooled estimate of the proportion \(p\).
A Z-score near 0 suggests no significant difference, while a larger or smaller Z-score might indicate a statistically significant difference.
Proportions Comparison
Proportions comparison is about evaluating the differences between two ratios or percentages from independent samples.
For example, this exercise compared the proportion of Republicans and Democrats citing the economy as an important issue.
We can use statistical tests to ascertain if observed differences could occur by chance or if they are statistically significant.
By using proportions comparison, we compute a metric like the difference \(p_1 - p_2\) and then analyze this difference through confidence intervals and hypothesis tests.
If the confidence interval for the difference doesn't contain zero, we conclude that there is a statistically significant difference between the proportions.
In this exercise, the confidence interval of \((-0.03, 0.01)\) contains zero, implying no significant difference.
Large-sample Estimation
Large-sample estimation refers to approximations used when the sample size is sufficiently large.
In statistical analysis, larger samples provide more reliable estimations of population parameters because the sample better represents the population.
This exercise uses large-sample methods to compare the proportions of Republicans and Democrats regarding economic issues.
The formulae for pooled proportions, standard errors, and resultant Z-scores rely on large sample approximations to the normal distribution.
For two groups with sizes \(n_1 = 995\) and \(n_2 = 1094\), it's considered a large enough sample for the Central Limit Theorem to apply.
This makes the Z-test a valid approximation for testing differences in proportions.
  • Pooled proportion: \(p = \frac{p_{1}n_{1} + p_{2}n_{2}}{n_{1}+n_{2}}\)
These methods rely heavily on having enough observations to ensure the normal distribution of the sample mean.

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Most popular questions from this chapter

The average American has become accustomed to eating away from home, especially at fast-food restaurants. Partly as a result of this fast-food habit, the per-capita consumption of cheese (the main ingredient in pizza) and nondiet soft drinks has risen dramatically from a decade ago. A study in American Demographics reports that the average American consumes 25.7 pounds of cheese and drinks 40 gallons (or approximately 645 8-ounce servings) of nondiet soft drinks per year. \({ }^{17}\) To test the accuracy of these reported averages, a random sample of 40 consumers is selected, and these summary statistics are recorded: $$\begin{array}{lcc} & \text { Cheese (lb/yr) } & \text { Soft Drinks (gal/yr) } \\\\\hline \text { Sample Mean } & 28.1 & 39.2 \\\\\text { Sample Standard Deviation } & 3.8 & 4.5\end{array}$$ Use your knowledge of statistical estimation to estimate the average per- capita annual consumption for these two products. Does this sample cause you to support or to question the accuracy of the reported averages? Explain.

Refer to Exercise 8.5 . What effect does an increased sample size have on the margin of error?

Red Meat, continued Refer to Exercise \(8.75 .\) The researcher selects two groups of 400 subjects each and collects the following sample information on the annual beef consumption now and 10 years ago: $$\begin{array}{lll} & \text { Ten Years Ago } & \text { This Year } \\\\\hline \text { Sample Mean } & 73 & 63 \\\\\text { Sample Standard Deviation } & 25 & 28\end{array}$$ a. The researcher would like to show that per-capita beef consumption has decreased in the last 10 years, so she needs to show that the difference in the averages is greater than \(0 .\) Find a \(99 \%\) lower confidence bound for the difference in the average per-capita beef consumptions for the two groups. b. What conclusions can the researcher draw using the confidence bound from part a?

Refer to Exercise \(8.3 .\) What effect does a larger population variance have on the margin of error?

A quality-control engineer wants to estimate the fraction of defectives in a large lot of film cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a \(95 \%\) confidence interval?

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